A uniform spherical shell of mass M = 9.30 kg and radius R = 0.470 m can rotate

Question

A uniform spherical shell of mass M = 9.30 kg and radius R = 0.470 m can rotate about a vertical axis on frictionless bearings (see the figure). A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 0.0540 kg.m^2 and radius r = 0.0950 m, and is attached to a small object of mass m = 3.00 kg. There is no friction on the pulleys axle; the cord does not slip on the pulley. What is the speed of the object when it has fallen a distance 1.40 m after being released from rest? Use energy considerations.

Explanation / Answer

The sum of the kinetic energies in the 2 rotating items will equal the loss in PE of the mass - the KE of the mass:

Is = (2/3)*9.3*0.47² =1.369 kgm²
Ip = 0.054 kgm²

Energy:
½Is*ws² + ½Ip*wp² = 3.0*g*1.4 - ½*3.0*v²

But ws= v/0.470 and wp = v/0.095, so

½*1.369*(v/0.47)² + ½*0.054*(v/0.095)² = 3.0*g*1.4 - ½*3.0*v²

(3.0986 + 2.991 + 1.5)v² = 41.16

                                   v² = 5.423

                                   v = 2.328 m/s

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